Question

. Spherically symmetric charge distribution. Fig. 3-15 shows a spherical distribution of charge of radius R. The charge density  at any points depends only on the distance of the point from the center and not on the direction, a condition called spherical symmetry. Find an expression for E for points (a) outside and (b) inside the charge distribution. Note that the object in Fig. 3-15 cannot be a conductor. 

Answer 1

Solution.

$\rho _0$ ;

$R;$

$r;$

$a) E_{out}=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_{tot}}{R^2};$

$q=\int\rho_0dV=\rho_0(\dfrac{4}{3}\pi R^3);$

$E_{out}=\dfrac{1}{4\pi \epsilon_0}\dfrac{\rho_0(\dfrac{4}{3}\pi R^3)}{R^2}=\dfrac{\rho_0 R}{3\epsilon_0};$

$b)$ $E_{in}=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_{within}}{r^2};$

$q=\int\rho_0dV=\rho_0(\dfrac{4}{3}\pi r^3);$

$E_{in}=\dfrac{1}{4\pi \epsilon_0}\dfrac{\rho_0(\dfrac{4}{3}\pi r^3)}{r^2}=\dfrac{\rho_0 r}{3\epsilon_0};$

Answer: $a)E_{out}=\dfrac{\rho_0 R}{3\epsilon_0};$

$b)E_{in}=\dfrac{\rho_0 r}{3\epsilon_0}.$